CBSE Class 10 Mathematics Areas Related to Circles Notes Set 04

Introduction

Many objects that we come across in daily life are related to the circular shape in some form or the other. So the problems for finding perimeters and areas related to circular figures are of great practical importance. In this chapter we learn how to find the perimeter (circumference) and area of circular figures and then apply this knowledge in finding the areas of two special 'parts' of a circular region known as sector and segment of a circle. We also learn how to find the areas of some combinations of plane figures involving circles or their parts.

Circle: Circle is the locus of a point which moves in such a manner that its distance from a fixed point O remains constant (the same). The fixed point is called the centre O and the constant distance OA is called its radius.

Chord

A line segment joining two points on a circle is called a chord of the circle. AB and CD are two chords in the figure above.

Diameter

A chord passing through the centre of the circle is called the diameter. Diameter is twice the radius. Here, AB is the diameter and the rest are chords.
Therefore \( d = 2r \), where d is the diameter and r is the radius.
The radius of a circle is the distance between any point on the circle and its centre. Therefore \( r = \frac{d}{2} \), where r is the radius and d is the diameter.
Secant: A line which intersects a circle at two points is called a secant of the circle.
Arc: A continuous piece of a circle is called an arc of the circle.

Circumference

The distance covered by travelling once around a circle is its perimeter or circumference. The circumference of a circle bears a constant ratio with its diameter, which is denoted by \( \pi \).\[ \frac{\text{circumference}}{\text{diameter}} = \pi \]Therefore, circumference \( = \pi \times \text{diameter} = \pi \times 2r \) [r is the radius of the circle and \( \pi \) is taken as \( \frac{22}{7} \) or 3.14].
Hence, circumference of the circle \( = 2\pi r \).
Circumference (perimeter) of a semicircle \( = \pi r + 2r \).

Area

Area of a circle \( = \pi r^2 \) or \( \frac{\pi d^2}{4} \) \( \left[\because r = \frac{d}{2}\right] \), where r is the radius and d is the diameter of the circle.
For a semicircle, area of a semicircle \( = \frac{1}{2} \pi r^2 \).
Area of a circular ring \( = \pi (R^2 - r^2) \), where R and r are the outer and inner radii of the circular ring respectively.

Knowledge Hub

1. In one revolution, the distance covered by a wheel is equal to the circumference of the wheel.
2. Number of revolutions in 1 minute by a wheel \( = \frac{\text{Distance covered in 1 minute}}{\text{Circumference}} \).

Example 1. If the circumference of a circle is 22 cm, then find the area of the circle.
Answer:
It is given that circumference \( = 22 \) cm. Circumference \( = 2\pi r \), so \( 2 \times \frac{22}{7} \times r = 22 \).
\( r = \frac{22 \times 7}{2 \times 22} = \frac{7}{2} \) cm.
Area of the circle \( = \pi r^2 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 38.5 \) cm².
Hence, the area of the circle is 38.5 cm².

Example 2. If the diameter of a wheel is 84 cm, then find the number of revolutions it must make to cover 792 metres.
Answer:
The diameter of the wheel is 84 cm, so its radius is \( \frac{84}{2} = 42 \) cm.
In one complete revolution, the distance covered by the wheel equals its circumference \( = 2\pi r = 2 \times \frac{22}{7} \times 42 = 264 \) cm.
Number of revolutions \( = \frac{\text{Distance covered by the wheel}}{\text{Circumference of the wheel}} = \frac{792\ \text{m}}{264\ \text{cm}} = \frac{79200\ \text{cm}}{264\ \text{cm}} = 300 \) (since 1 m = 100 cm).
Hence, the number of revolutions is 300.

Example 3. The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle.
Answer:
Let the radius of the circle be r cm. Then the diameter is 2r cm and the circumference is \( 2\pi r \) cm.
Given: circumference \( = \) diameter \( + 16.8 \Rightarrow 2\pi r = 2r + 16.8 \Rightarrow 2 \times \frac{22}{7} \times r = 2r + 16.8 \).
\( 44r = 7(2r + 16.8) \Rightarrow 44r = 14r + 117.6 \Rightarrow 30r = 117.6 \Rightarrow r = \frac{117.6}{30} = 3.92 \).
Therefore, the radius of the circle is 3.92 cm.

Example 4. A bucket is raised from a well with the help of a rope which is wound around a wheel of diameter 77 cm. Given that the bucket ascends in 1 min. 28 seconds with a uniform speed of 1.1 m/sec. Find the number of complete revolutions the wheel makes in raising the bucket.
Answer:
Time taken by the bucket to ascend \( = 1 \) min. 28 sec. \( = 88 \) sec. and the speed is 1.1 m/sec.
Distance covered by the bucket \( = \) length of the rope \( = 1.1 \times 88 = 96.8 \) m \( = 9680 \) cm.
Radius of the wheel \( = \frac{77}{2} \) cm, so the circumference of the wheel \( = 2\pi r = 2 \times \frac{22}{7} \times \frac{77}{2} = 242 \) cm.
Number of revolutions \( = \frac{\text{Length of the rope}}{\text{Circumference of the wheel}} = \frac{9680}{242} = 40 \).
Hence, the number of revolutions is 40.

Example 5. A racetrack is in the form of a ring whose inner circumference is 352 m, and the outer circumference is 396 m. Find the width of the track.
Answer:
Let R and r be the outer and inner radii of the ring respectively. Then \( 2\pi R = 396 \) and \( 2\pi r = 352 \).
Subtracting: \( 2 \times \frac{22}{7} \times (R - r) = 396 - 352 = 44 \Rightarrow R - r = \frac{44 \times 7}{44} = 7 \) m.
Width of the track \( = R - r = 7 \) m (here \( R = 63 \) m and \( r = 56 \) m).
Hence, the width of the track is 7 m.